Tom Throws a Ball: Horizontal & Vertical Velocity

[littledude565]

Homework Statement

Tom throws a ball of a cliff 45m nigh, with a velocity of 30 m/s. Take accleration due to gravity as 10m/s. (It took 4.5s to reach the bottom)

What was its horizontal and verticle velocitys when it hit the ground?
i used the first equation (v=u+at) and with substituion i got -42m/s for the verticle velocity. Is this correct, and how would i find the horizontal? thanks for the help

 

[LowlyPion]

Homework Statement

Tom throws a ball of a cliff 45m nigh, with a velocity of 30 m/s. Take accleration due to gravity as 10m/s. (It took 4.5s to reach the bottom)

What was its horizontal and verticle velocitys when it hit the ground?
i used the first equation (v=u+at) and with substituion i got -42m/s for the verticle velocity. Is this correct, and how would i find the horizontal? thanks for the help

Welcome to PF.

If your initial velocity was horizontal, then the time to fall is not given by the velocity relationship. That equation would be useful if you wanted to know how long it took to reach its height i.e when velocity goes to 0.

To find the time you need to use the relationship that relates distance and acceleration and time.

x = 1/2*g*t²

 

[littledude565]

the g stands for gravity right? if so would the answer for the verticle be 101.25?

 

[littledude565]

and the horizontal 30m/s as its a constant?

 

[LowlyPion]

the g stands for gravity right? if so would the answer for the verticle be 101.25?

Based on what?

You'll have to show where that comes from.

 

[littledude565]

in your equation x = 1/2*g*t²

So using that i did (assuming the G did stand for gravity). 1/2 * 10m/s*4.5s^2 and i got an answer of 101.25

 

[LowlyPion]

in your equation x = 1/2*g*t²

So using that i did (assuming the G did stand for gravity). 1/2 * 10m/s*4.5s^2 and i got an answer of 101.25

You've run out the corral without the saddle there.

The height is given as 45 m. Your time calculation as I said already is not based on the first equation you applied. 4.5 sec is just plain wrong.

So start again with the right equation and find the correct time and then you can figure it out.

 

[littledude565]

Ok using the s=1/2*(u+v)*t i did 45=15t, t=45/15 and so i got 3seconds. Is that correct?

 

[LowlyPion]

Ok using the s=1/2*(u+v)*t i did 45=15t, t=45/15 and so i got 3seconds. Is that correct?

3 seconds is a much better value to use.

 

[littledude565]

so what equation do i use now? :S

 

[littledude565]

ok so if the verticle velocity increases by 10 every second, would the answer then be 30m/s?

 

[LowlyPion]

ok so if the verticle velocity increases by 10 every second, would the answer then be 30m/s?

Yes the vertical is 30 m/s at impact. As is the horizontal incidentally.

 

[littledude565]

ah thank you very much. So would that mean the resultant of these would be 54.08? I got that using phythag sqrt 45^2+30^2 = 54.08 with an angle of 34

 

[LowlyPion]

ah thank you very much. So would that mean the resultant of these would be 54.08? I got that using phythag sqrt 45^2+30^2 = 54.08

Why do you think it's 45² again?

 

[littledude565]

http://img21.imageshack.us/img21/7229/83881376.png

 

[littledude565]

ohhh no it should be 45 it should also be 30! So with that now would the resultant be 42.2, with an angle of 45?

 

[littledude565]

http://img24.imageshack.us/img24/9337/71116538.png

 

[LowlyPion]

Yes. That would be better now.

 

[littledude565]

Ah yay, thanks for all your help